Question: Is ${522183}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {522183}= &&{5}\cdot100000+ \\&&{2}\cdot10000+ \\&&{2}\cdot1000+ \\&&{1}\cdot100+ \\&&{8}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {522183}= &&{5}(99999+1)+ \\&&{2}(9999+1)+ \\&&{2}(999+1)+ \\&&{1}(99+1)+ \\&&{8}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {522183}= &&\gray{5\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {5}+{2}+{2}+{1}+{8}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${522183}$ is divisible by $9$ if ${ 5}+{2}+{2}+{1}+{8}+{3}$ is divisible by $9$ Add the digits of ${522183}$ $ {5}+{2}+{2}+{1}+{8}+{3} = {21} $ If ${21}$ is divisible by $9$ , then ${522183}$ must also be divisible by $9$ ${21}$ is not divisible by $9$, therefore ${522183}$ must not be divisible by $9$.